74 items are filed under this topic.
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Dividing the tips |
2020-01-02 |
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From Pat: trying to figure out division of tips I have 3 full time employees 40 hrs a week and 1 part time worker 12 hours a week what percentages do i give them to give them appropriate tips :-) Answered by Harley Weston. |
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Liza works for a call center |
2019-10-31 |
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From Timothy: Liza knows that in the long term, she has a 65% chance of making a sale when calling on customers. One morning, he makes Six calls.
a. What is her probability of making three calls?
b. What is her probability of making sales fewer than three? Answered by Penny Nom. |
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A binomial probability question |
2019-10-08 |
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From venni: The Medassist Pharmaceutical Company receives large shipments of aspirin
tablets and uses this acceptance sampling plan: Randomly select and test 24
tablets, then accept the whole batch if there is only one or none that doesn’t
meet the required specifications. If a particular shipment of thousands of
aspirin tablets actually has a 4% rate of defects, what is the probability that
this whole shipment will be accepted? Answered by Penny Nom. |
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The binomial distribution |
2019-10-01 |
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From PY: An on-line game called ‘Shop Quiz’ is held by an e-commerce platform, from Monday to Friday
every week. It consists of 8 multiple choice questions (MCQ) and each question has four options (A, B, C, D). Only one option is the correct answer. People who are able to correctly answer all 8
questions are winners and will be awarded a number of on-line shopping credits.
Let X represent the number of questions that a person can answer correctly in a ‘Shop Quiz’.
1) Explain why Binomial distribution might NOT be a suitable distribution for the random variable
X.
Mr. Saul likes playing the quiz, however, he is afraid that he might not have the necessary knowledge to answer the quiz questions. (The quiz questions cover a variety of topics including science, history, entertainment, sports and geography, etc.) Therefore, he tries to win the game by simply guessing the answers to each question. Answered by Penny Nom. |
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Sampling distribution |
2017-10-16 |
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From Esther: I need 4digits combination of 235689 and without replacement. Thanks Answered by Penny Nom. |
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A normal distribution problem |
2014-04-19 |
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From Melanie: This is the question:
The lifetime of a certain type of car tire are normally distributed. The mean lifetime of a car tire is 40,000 miles with a standard deviation of 5,000 miles. Consider a sample of 10,000 tires. A) How many tires would you expect to last between 35,000 and 45,000 miles? b) How many tires would you expect to last between 30,000 and 40,000 miles? c) How many tires would you expect to last less than 40,000 miles? d)How many tires would you expect to last more than 50,000 miles? e) How many tires would you expect to last less then 25,000 miles? f) What tires would you want on your car and explain your reasoning
Not at all sure that we've done any of this correctly and not sure how to determine how many tires will last less than 25,000 miles.
Any help is appreciated. Answered by Penny Nom. |
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A normal distribution problem |
2012-05-13 |
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From Alysia: The scores on a test taken by 1000 students are normally distributed with a mean of 66 and standard of deviation of 12. If the college wishes only the top 8% of people to get an A, what would the cutoff score be for the A's? Answered by Penny Nom. |
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A normal distribution problem |
2010-12-02 |
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From Racquel: I am stuck on this question. I am not sure if using the z score will help
me get the answer I need. Here is the question?
The average length of time per week that students at this university spend
on homework is normally distributed with a mean of 18 hours and a standard
deviation of 3 hours. If Diane spends more time on homework each week than
75% of students, what is the minimum time she must spend? Answered by Penny Nom. |
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Sample size |
2010-03-29 |
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From Rae: What sample size was needed to obtain an error range of 2% if the following statement was made? "75% of the workers support the proposed benefit package. These results are considered accurate to within + or - 2%, 18 out of 20 times. This seems like a straight forward question but I'm getting it wrong. Could you please help me out even just the set up would be appreciated so I can see if that's where I'm going wrong. Thanks Answered by Harley Weston. |
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The binomial distribution |
2010-02-27 |
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From Jessica: A small mobile phone retailer has found that one of their phones has a 12% probability of being faulty and a replacement having to be provided for the customer. They have just
received a trial order for 10 phones from their biggest customer who will take their business elsewhere if 20% or more items are faulty.
i) what is the probability that they will lose their biggest customer? Answered by Penny Nom. |
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Normal Distribution |
2009-09-02 |
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From Nikita: Scores on a college exam are known to be normally distributed with a standard deviation of 20. If the top 3% have scores in excess of 200, what is the mean score? Answered by Robert Dawson. |
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Z-score |
2009-03-25 |
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From Barb: I am having trouble finding information on a z score and the conversion to the number of standard deviations a z value can be away from the mean.
What exactly does that mean and what am I looking for?
Help Please. Answered by Harley Weston. |
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A normal distribution problem |
2009-03-21 |
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From EDGAR: to qualify for security officers training recruits are tested for stress tolerance. The scores are normally distributed with mean of 62 and a standard deviation of 8.
a.) If only the top 15% of recruits are selected, find the cutoff score
b.) If a candidate is rendomly selected, what is the probability that his or her socre is at least 55? Answered by Harley Weston. |
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Percentiles |
2009-03-21 |
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From Shawn: For a normal distribution of u=654.00 and o=138.00.
What is the percentile rank for X=426? Answered by Harley Weston. |
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A normal distribution problem |
2009-03-13 |
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From jude: Regarding the time it takes for an oil change has a normal distribution with a mean of 17.8 minutes and std. deviation of 5.2 minutes. A free oil change will be given to any customer that must wait beyond the guaranteed time. If they don't want to give more than 1% of its customers free oil changes how long should the guarantee be (to the nearest minute). Thank you. Answered by Robert Dawson. |
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The distribution of sample sums |
2008-11-21 |
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From Mark: For large samples, the sample sum (Σ x) has an approximately normal distribution.
The mean of the sample sum is n*μ and standard deviation is (σ*√n). The distribution of savings per account for savings and loan institution has a mean equal to $750 and a standard deviation equal to $25. For a sample of 50 such accounts, find the probability that the sum in the 50 accounts exceeds $38,000. Answered by Penny Nom. |
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Sigma in a normal distribution |
2008-11-18 |
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From Justin: Suppose the random variable Y can be described by a normal curve with
Mu=40. For what value of the standard deviation is
P(20 less than or equal to Y less than or equal to 60) = 0.50
-Justin Answered by Harley Weston. |
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A normal distribution problem |
2008-11-18 |
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From Mark: Final Averages are typically approximately normally distributed with a mean of 72 and a standard deviation of 12.5.
your professor says that the top 8% of the class will receieve an A, the next 20%, a B, the next 42%, a C and the bottom 12%, an F.
a. What average must you exceed to obtain an A?
b. What Average must you exceed to receieve a grade better than a C?
c. What average must you obtain to pass the course? (you'll need a D or better) Answered by Harley Weston. |
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Trinomial Distribution |
2008-10-22 |
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From Darya: show that (x^2-x+3)(2x^2-3x-9) = Ax^4+Bx^3+C where A,B and C are constants to be found. Answered by Janice Cotcher. |
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A multiple choice exam |
2008-09-13 |
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From Phalange: A multiple choice exam consists of 12 questions, each having 5 possible answers. To pass, you must answer at least 9 out of 12 questions correctly. What is the probability of passing if:
a. You go into the exam without knowing a thing, and have to resort to pure guessing?
b. You have studied enough so that on each question, 3 choices can be eliminated. But then you have to make a pure guess between the remaining 2 choices.
c. You have studied enough so that you know for sure the correct answer on 2 questions. For the remaining 10 questions you have to resort to pure guessing. Answered by Harley Weston. |
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z-score |
2008-07-02 |
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From Candace: If a normal-shaped distribution has a mean of 80 and a standard deviation of 15 what is the z-score for M=84 for a sample of n=25 scores. Is this sample mean in the middle 95% of the distribution? Answered by Harley Weston. |
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The distribution of sample means |
2008-07-02 |
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From crystal: A population forms a normal distribution with a mean of 75 and a standard deviation of 20. How do I sketch the distribution of sample means for samples of n= 100? Answered by Harley Weston. |
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The mean and variance |
2008-06-05 |
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From Donny: An investment will be worth $1,000, $2,000, or $5,000 at the end of the
year. The probabilities of these values are .25, .60, and .15, respectively.
Determine the mean and variance of the worth of the investment. Answered by Harley Weston. |
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A coin is tossed 30 times |
2008-04-13 |
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From seulki: coin tossed 30 times, what is the probability that the heads show up fewer than 17 times? Answered by Harley Weston. |
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A normal distribution problem |
2008-03-29 |
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From Lorie: 3. The amount of annual snowfall in a certain mountain range is normally distributed with a mean of 109 inches, and a standard deviation of 10 inches.
a. What is the probability that the mean annual snowfall during 40 randomly picked years will exceed 111.8 inches? Answered by Harley Weston. |
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Probability and Merta trains |
2008-03-27 |
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From Lorie: 4. Merta claims that 74% of its trains are on time.
a. Find the probability that among the 60 trains, 38 or fewer arrived on time. Answered by Harley Weston. |
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A probability distribution |
2008-03-23 |
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From Lorie: The probabilities that a batch of 4 computers will contain 1, 2, 3, and 4 defective computers are 0.6274, 0.3102, 0.0575, and 0.001 respectively.
a. Set up a probability distribution to describe this situation. Answered by Penny Nom. |
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Expected value and standard deviation |
2008-03-19 |
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From Patrick: Heres a question i cant figure out:
a small airline company has only three flights per day. The number of
delayed flights per day is regarded as a random variable, and I'm
supposed to calculate the expected
value and standard deviation of the number of delays.
the probability distribution looks like:
No. of Delays: 0 1 2 3
Prob. of delay: 05. 0.3 .1 .1 Answered by Harley Weston. |
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A pre-employment evaluation |
2008-02-06 |
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From lisa: An employer gives a pre-employment evaluation to a large group of applicants.
The scores are normally distributed with a mean of 154 and a standard
deviation of 21. The employer wants to interview only those applicants
who score in the top 15%. What should the cut off score be for the interviews? Answered by Harley Weston. |
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p-values and t-distributions |
2007-11-13 |
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From Don: I have a t-test statistic of 1.28 and a degree of freedom of 6.
I know the p-value is .248 I cannot figure out how to calculate that value.
I have used software to get it but I want to know how to calculate it using
the t-distribution table. Answered by Harley Weston. |
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A normal distribution problem |
2007-11-11 |
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From Jenny: I am a part-time student so that i have no time to ask the lecturer. moreover the book which i borrowed from state library don't have any answer. but i have already done with most of the question. but these three question which i attached is really confusing me. i am very glad that you help me. Answered by Harley Weston. |
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A multiple choice exam |
2007-10-21 |
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From jon: a student is taking a multiple choice exam in which each question has four choices.
assuming that she has no knowledge of the correct answers to any of the questions, she
has decided on a strategy in which she will place four balls (marked A, B, C, D) into a box. she randomly
selects one ball for each question and replaced the ball in the box. the marking will determine her answer to the question.
there are five multiple choice questions on the exam. what's the probability that she will ... Answered by Stephen La Rocque and Harley Weston. |
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The normal approximation to the binomial |
2007-09-30 |
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From m.j.: Slot Machines The probability of winning on a slot machine is 5%. If a person plays the machine 500 times, find the probability of winning 30 times. Use the normal approximation to the binomial distribution. Answered by Harley Weston. |
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A normal distribution problem |
2007-09-27 |
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From m.j.: Car Loan Rates The national average for a new car loan was 8.28%. If the rate is normally distributed with a standard deviation of 3.5%, find these probabilities.
a. One can receive a rate less than 9%.
b. One can receive a rate less than 8%. Answered by Harley Weston. |
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The weights of packages are normally distributed |
2007-09-23 |
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From alan: the weight of a packet of sweets produced in a factory are normally distributed.
the mean weight is 100g
the standard deviation is 2g
all packets weighing less than 99g and more then 105g are rejected
what proportion are rejected Answered by Penny Nom. |
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Standard deviation |
2007-09-01 |
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From Pat: I have looked over your math question and mines is not there. for a binomiial distribution with parmeter n and p, the mean is np and the standard deviation is np(1-p). I understand how to get the np but not understanding how to get the (1-p). For example np=(40)(0.48)(1-0.48)=3.160. How can I get the (1-0.48) answer please Answered by Harley Weston. |
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A statistics example |
2007-08-02 |
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From Claudia: A particular employee arrives to work some time between 8:00 am - 8:30 am. Based on past experience the Company has determined that the employee is equally likely to arrive at any time between 8:00 am - 8:30 am.
On average, what time does the employee arrive?
What is the standard deviation of the time at which the employee arrives?
Find the probability that the employee arrives exactly at 8:12 am?
Find the probability that the employee arrives between 8:20 am - 8:25 am? Answered by Har. |
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Standard Deviation |
2007-06-13 |
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From Adrian: If you are told that the mean salary of a certain group of workers is $30,000 with a standard deviation of $4000, what proportion of workers earn over $38,000? What proportion of workers earn less than $18,000? Assume the distribution of wages is normal. Answered by Stephen La Rocque. |
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Normal distribution |
2007-05-24 |
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From Paula: Consider a data set that is normally distributed.
The mean of the data set is equal to 10,000.
a.) Suppose that, for this data set, 10,625 has a
"z-value" = 2.5. Solve for the standard deviation of
the data set.
b.) Solve for the "z-value" of 9,900. Answered by Penny Nom. |
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Find the sample size needed |
2007-05-13 |
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From Mini: Find the sample size needed to be 98% confident thata marketing survey on the proportion of shoppers who use the internet for holiday shopping is accurate within a margin of error of 0.02. Assume that the conditions for a binomial distribution are met, and that a current estimate for a sample proportion does not exist. Answered by Penny Nom. |
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Permutations, probability and standard normals |
2007-05-09 |
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From Katrina: I have a few problems that i seem to be stuck on or can not start. Can you please help me ?
1) There are 20 people on an event planning committee. How many differnt ways can a
chairperson and assistant to the chairperson be selected?
2) An unprepared student makes random guesses for 10 true or false questions on a quiz.
Find the probability that the student passes the quiz by guessing 7 of the questions
correctly.
3) The heights of 18 year old men are normally distributed with a mean of 68 inches and
a standard deviation of 3 inches. If a random sample of 25 18-year old men is selected
what is the probability that the mean height is between 68.5 and 72 inches? Answered by Penny Nom. |
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A normal distributiion question |
2007-04-20 |
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From Erika: The amount of time required for a certain type of automobile transmission repair at a service garage is normally distributed with the mean = 45 minutes and the standard deviation =8.0 minutes. The service manager plans to have work begin on the transmission of a customer’s car 10 minutes after the car is dropped off, and he tells the customer that the car will be ready within one hour total time. What is the probability that he will be wrong? Illustrate the proportion of area under the normal curve which is relevant in this case.
What is the required working time allotment such that there is a 75 percent chance that the transmission repair will be completed withing that time? Illustrate the proportion of area that is relevant. Answered by Penny Nom. |
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A hypothesis test |
2007-04-09 |
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From Katrina: I have already tried to do this problem but im having a very had time with
it. Can you please help me.
Glamour Magazine sponsored a survey of 2500 prospective brides and
found that 60% of them spent less than $750 on their wedding gown.
Use a 0.01 significance level to test the claim that less than 62% of brides
spend less than $750 on their wedding gown. How are the results
affected if it is learned that the responses were obtained from magazine
readers who decided to respond to the survey through an Internet Web
site? Answered by Penny Nom. |
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Acceptance sampling |
2007-03-29 |
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From Katrina: The Medassist Pharmaceutical Company recieves large shipments of
asprin tablets and uses this acceptance sampling plan: Randomly select
and test 24 tablets, then accept the whole batch if there is only one or
none that doesnt meet the required specifications. If a particular shipment
of thousands of asprin tablets actually has a 4% rate of defects, what is
the probability that this whole shipment will be accepted? Answered by Penny Nom. |
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What is the expected number of ripe and ready to eat watermelons |
2006-11-29 |
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From James: An agricultural cooperative claims 95 percent of the watermelons shipped out are ripe and ready to eat. If 20 watermelons are shipped out , what is the probability that the number of watermelon that are ripe and ready to is (i) exactly 14 (ii) more than 18 (iii) of the 20 watermelons that are shipped what is the expected number of ripe and ready to eat watermelons Answered by Penny Nom. |
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Sally plays roulette |
2006-11-21 |
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From Sherry: Sally plans to bet 100 bets of $1.00 each on red roulette. the probability of the ball landing on red is 18/38 what is the probability that Sally will win at least half the time Answered by Penny Nom. |
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The coefficient of variation |
2006-05-20 |
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From Glenn: What is the correct formula for coefficient of variation for a binomial distribution? Answered by Penny Nom. |
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A manufacturer of cotton pins |
2006-03-20 |
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From Nirmal: A manufacturer of cotton pins knows that 5% of his products are defective. If he sells cotton pins in boxes of 100 and guarantees that not more than 10 pins will be defective, what is the approximate probability that a box will have the guaranteed quantity? Answered by Penny Nom. |
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Another normal distribution problem |
2006-02-18 |
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From Mary: Assume that blood pressure readings are normally distributed with a mean of 120 and standard deviation of 8. A researcher wishes to select people for a study but wants to exclude the top and bottom 10 percent. What would be the upper & lower readings to qualify people to participate in the study? Answered by Penny Nom. |
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A normal distribution problem |
2006-02-15 |
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From Mary: In a certain normal distribution, find the mean when the standard deviation is 5 and 5.48% of the area lies to the left of 78. Answered by Penny Nom. |
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A binomial distribution exercise |
2006-01-21 |
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From Belinda: In a survey of 15 manufacturing firms, the number of firms that use LIFO (a last-in first-out accounting procedure for inventory) is a binomial random variable x with n=15 and p=0.2.
a) What is the probability that five or fewer firms will be found to use LIFO? Is it unlikely that more than 10 firms will be found to use LIFO? Comment. Answered by Penny Nom. |
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A confidence interval |
2006-01-21 |
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From Jonathan:
I am attempting to calculate how my confidence interval will widen at the 95% confidence level if my response universe increases from 100 to 150 or to 200.
There is a universe of 54,000. I take a 5% sample for a test universe of 2,700
If my "yes" universe is 100, at the 95% confidence level, what is my +/- range? (i.e +/- 3? +/-5?)
Historically, 6.6% of the 2,700 you say "yes". I am trying to determine how the confidence interval would change if the number of "yes" responders increased to 150 or to 200.
Answered by Penny Nom. |
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A sample size estimation |
2005-12-03 |
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From Ivonne: I have to do a research about the behavior of library users. We are going to apply a survey to a population of 1280 students (Management an Economics students) but of course we have to do it to a sample....I need to know the size of my sample. Answered by Penny Nom. |
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The distributive property |
2005-11-12 |
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From Jennifer: I am trying to help my daughter . How do you name the property for 6(4w - 3z)=(6 x 4w)-(6 x 3z) Answered by Penny Nom. |
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A normal distribution problem |
2005-08-08 |
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From Brad: The life of a toy is normally distributed. Suppose 92.51% of the items lives exceeding 2,160 hours and 3.92% have lives exceeding 17,040 hours. Find the mean and the standard deviation. Answered by Penny Nom. |
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Computing confidence intervals |
2004-11-26 |
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From Christie: I was given a question with N=100, sample proportion is 0.1- compute the 95% confidence interval for P? I have tried this several ways but do not know how to do without means, standard deviations, standard error of the mean? I asked my teacher and she said I have all the info I need. Can you help???? Answered by Penny Nom. |
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A probability question |
2004-08-15 |
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From Gary: In a law school class, the entering students averaged about 160 on the last
LSAT; the standard deviation was about 8. The historgram of the lLSAT
scores follwed the normal curve reasonable well.
Q. About what percentage of the class scored below 166?
Q. One student was 0.5 above average on the last LSAT, about what percentage
of the students had lower scores than he did? Answered by Penny Nom. |
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The p-value |
2004-07-25 |
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From Kathy: Ms. Lisa Monnin is the budget director for the New Process Company. She would like to compare the daily travel expenses for the sales staff and the audit staff. She collected the following sample information.
At the .10 significance level, can she conclude that the mean daily expenses are greater for the sales staff than the audit staff? What is the p-value?
Having problems finding the p-value & unsure of the formula.
Kathy Answered by Penny Nom. |
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Replacement times for TV sets? |
2004-03-31 |
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From Barb: Replacement times for TV sets are normally distributed with a mean of 8.2 years and a standard deviation of 1.1 years. Estimate the probability that for 250 randomly selected TV sets, at least 15 of them have replacement times greater than 10.0 years.e Answered by Andrei Volodin and Penny Nom. |
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Binomial distribution |
2003-12-17 |
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From Lesley: my daughter is having difficulty with the following formula
P(X=x) = ( n over x) px (1-p) n-x
The teacher has given them the formula but not taught them how to apply it or understand it. Answered by Penny Nom. |
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Standard Deviation |
2003-10-07 |
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From Rebecca:
I have a task to complete, which is to calculate the mean and standard deviation of something. I have done this but am then asked to write a short explanation of my findings.
I know what the mean is about, and I thought I knew what the standard deviation meant too - shows the variation from the mean. However, on a task I completed earlier the feedback I got said 'you need to tell us that it is talking about the middle 66% of the data' - that has thrown me, I don't understand that. Can anyone help me get my head round this??? Answered by Penny Nom. |
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A Normal probability problem |
2002-12-03 |
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From A student: The height of married men is approximately normal with mean 70 and standard deviation 3. The height of married women is approximately normal with mean 65 and standard deviation 2.5. What is the probability that a random married woman is taller than a random married man? Answered by Andrei Volodin. |
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Day care |
2002-05-13 |
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From Sonam: In many familes, both parents work. as a result, there is increasing need for day care. data was collected; and in one year in Canada, approximately 32% of children aged 0 to 11 years were in day care for at least 20h per week. (a) what is the probability, in a random poll of 60 children form the age of 0 to 11, that more than 15 children are in day care at least 20 h per week? nearest tenth of one %
ANSWER: P(children are in daycare at least 20h)= 60/60C14 = to the answer (b) what is the probability, in a random pool of 60 children that fewer than 20 are in day care at least 20 h per week?
ANSWER: P= 20/60= 33.3% stay in day care for 20h per week, I dont know if these answers are right please help me out. Answered by Andrei Volodin. |
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Sampling distributions |
2002-02-18 |
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From A student:
- given: n = 40, standard deviation is not known, population of individual observations not normal. does the central limit theorem apply in this case? why or why not?
- for an estimation problem, list two ways of reducing the magnitude of sampling error?
- What will happen to the magnitude of sampling error if the confidence level is raised all other things remaining the same? justify your answer?
Answered by Harley Weston. |
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Normal distribution |
2002-01-21 |
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From Danielle: A teacher gave a test on which the students' marks were normally distributed, but the results were pathetic. The mean was 52% and the standart deviation was 12%. The teacher decided that the top 10% of the students should get A's, the next 20% should get B's, the next 40% should get C's, the next 20% should get D's, and the bottom 10% should get F's. To the nearest percent, what are the cutoff marks that will result in an A, B, C, D, and F? Answered by Penny Nom. |
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Cinderella clothes |
2001-12-11 |
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From A student: If cinderella clothes, inc. has determined that 0.5% of all incoming phone calls involve complaints, what is the probability that in 200 incoming calls there are more than one complaint? Answered by Andrei Volodin. |
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A confidence interval |
2001-06-28 |
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From Murray: An investigator wants to find out of there are any difference in "skills" between full and part time students. Records show the following:
Student Mean Score Std Dev Number
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Full time 83 12 45
Part time 70 15 55
Compute a 95% confidence interval for the difference in mean scores. Answered by Andrei Volodin. |
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Banana yogurt |
2000-11-03 |
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From James: A grocery store has 100 cartons of banana yogurt in stock.Each carton contains 12 cup of banana yogurts.The probability that a cup has fewer than 20 banana chunks in it is 10 %. So,What is the probability that between 15 and 25 (inclusive) cartons out of the 100 cartons have exactly 3 cups with fewer than 20 banana chunks? Answered by Harley Weston. |
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Independent tests |
2000-10-07 |
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From A student: If the false-positive rate of each test in a battery of tests is 0.05, how many independent tests can be included in the battery if we want the probability of obtaining at least one false-positive result to be at most 0.2? Answered by Harley Weston. |
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Central Limit Theorem and Law of Large Numbers |
2000-06-26 |
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From Jonathan Yam: The Central limit Theorem states that when sample size tends to infinity, the sample mean will be normally distributed. The Law of Large Number states that when sample size tends to infinity, the sample mean equals to population mean. Is the two statements contradictory? Answered by Paul Betts and Harley Weston. |
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Standard Deviation |
1999-07-11 |
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From Anthony Fama: I have seen several answers to this question: If one standard deviation represents 68% of the population, what does two, three, four and five sigma [std deviation] represent? As stated, I have seen several different answers and thus, the impetus for my question. Answered by Harley Weston. |
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The Central Limit Theorem |
1997-04-21 |
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From Donna Hall: A skeptic gives the following argument to show that there must be a flaw in the central limit theorem: We know that the sum of independent Poisson random variables follows a Poisson distribution with aparameter that is the sum of the parameters of the summands. In particular, if n independentPoisson random variables, each with parameter 1/n, are summed, the sum has a Poisson distributionwith parameter 1. The central limit theoren says the sum tends to a normal distribution, butPoisson distribution with parameter 1 is not normal. What do you think of this argument? Answered by Neal Madras. |
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The normal distribution. |
1997-03-21 |
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From Donna D.Hall: I am looking for a proof for the normal distribution. I suppose "proof" was not a good choice of words. What I am looking for is a way to "derive" the normal distribution in simple terms so that the most average teenager can see the logic. Can you help me? Answered by Harley Weston. |
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What is the variance of the difference of two binomials? |
1996-02-08 |
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From Chris Johnson: My co-workers and I have come up with different estimates of the z-statistic, and are in particular disagreement over the calculation of Variance for this problem. I am trying to find out, with a five percent level of significance, whether the new form yields a higher rate of return than the old form. Any thoughts, comments, or solutions? Answered by Harley Weston. |
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